Math Puzzler #3

There are six Dudeney numbers, positive integers whose decimal digit sum cubed is equal to the original number. they are
amath 1=1^3=(1)^3 endmath,
amath 512=8^3=(5+1+2)^3 endmath,
amath 4913=17^3=(4+9+1+3)^3 endmath,
amath 5832=18^3=(5+8+3+2)^3 endmath,
amath 17576=26^3=(1+7+5+7+6)^3 endmath,
and amath 19683=27^3=(1+9+6+8+3)^3 amath.
  1. Prove there are no larger Dudeney numbers.
  2. Find all numbers where the fourth power of the digit sum is equal to the number itself.
Hint: A computer may be useful to check cases, but it is possible to do this problem by hand, albeit with a bit of paper.
Solution 1 after the jump, Solution 2 next week.
  1. I've done this all by hand, and I'll write it out here(save a bunch of space consuming checks at the end) then post python code I used to check the solutions more efficiently.
    First, let's show that sufficiently large numbers cannot be Dudeney (Thanks to Hostile Fork for the idea). Note that
    amath DigitSum ( n ) <= 9 * DigitCount ( n ) endmath and amath DigitCount ( n ) < log_10 ( n ) + 1 endmath so amath DigitSum ( n ) < 9*log_10 ( n ) + 9 endmath. amath n^3 endmath is a Dudeney number if and only if amath n = DigitSum ( n^3 ) endmath. This means amath n < 27*log_10 ( n ) + 9 endmath. By hand, it's clear that amath 10^2=100 endmath is an upper bound for amath n endmath. Solving the inequity above with a computer shows that amath n < 57 endmath. Either way,now we can check each possible n to see if it is a Dudeney number. 19683 is indeed the largest Dudeney number. Here's my code.
  2. If you looked at my code above, the solution here should be obvious. We know we want amath n = DigitSum ( n^4 ) endmath, so amath n < 36*log_10 ( n ) + 9 endmath.This gives us amath n<77 endmath. Testing cases gives us that 1, 2401, 234256, 390625, 614656, and 1679616 are the fourth powers which satisfy our condition. My code.

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